August 28th, 2016, 05:16 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 
You posted 8 problems, but showed no attempts at solutions. Are you trying to get your homework done here? Concerning this problem, since posted in "Geometry section", is it assumed that no trigonometry is allowed? 
August 28th, 2016, 06:20 PM  #3  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Quote:
Last edited by skipjack; October 28th, 2018 at 09:01 PM.  
August 28th, 2016, 07:25 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025  You mean "only works when point M is at center of BC", right?
Last edited by skipjack; October 28th, 2018 at 09:01 PM. 
August 29th, 2016, 06:54 AM  #5 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Yes, that is what I meant. Even though I was born and educated in the U.S., in school I took English as a foreign language.
Last edited by skipjack; October 28th, 2018 at 09:02 PM. 
August 29th, 2016, 08:06 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025  
October 28th, 2018, 02:09 PM  #7 
Newbie Joined: Oct 2018 From: usa Posts: 1 Thanks: 0 
Let I be the point of intersection of (AN) and (HC) angle AMN = angle (HBC) ...... (1) BM/AM = BH/HM THEN 2(BM/AM) = BH/(HM/2) OR BC/AM = BH/MN SO BC/BH = AM/MN ........ (2) (1) AND (2) MEANS THAT THE 2 TRIANGLES CBH AND AMN ARE SIMILAR AND ANGLE NAM = ANGLE HCB THEREFORE THE QUADRILATERAL AIMC IS CYCLIC AND x = 90° Last edited by skipjack; October 28th, 2018 at 11:15 PM. 

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